Suppose that $h_{1},\cdots,h_{n}\in \mathbb{R}^{p}$ are $n$ constant vectors, $g(\lambda)$ is a $p-$demensional function of a vector $\lambda \in \mathbb{R}^{p}$ defined as below: \begin{equation} g(\lambda) = \sum_{i=1}^{n}\frac{h_{i}}{1+\lambda^{\prime}h_{i}}, \end{equation} which has negative-definite derivative matrix $\frac{\partial g}{\partial \lambda}(\lambda)=-\sum_{i=1}^{n}\frac{h_{i}h_{i}^{\prime}}{(1+\lambda^{\prime}h_{i})^{2}}<0$ ($\lambda^{\prime}$ denotes the transpose of the vector $\lambda$).
My question is whether $g(\lambda)=0$ has roots within the convex set $\{\lambda| 1+\lambda^{\prime}hi>0 \text{ for } i=1,\cdots,n\}$. And if there is a unique root.
Thank you a lot for your help!
Here is an answer for a specific case. Suppose that $h_1,\ldots h_n$ are linearly independent. Set $c_i(\lambda) = 1/(1+\lambda'h_i)$ for each $i$. Now if $g(\lambda_0)= 0$ in the region $1+\lambda'h_i>0$, then $c_i>0$ for each $i$ and so $$ g(\lambda_0) = \sum_{i=1}^nc_i(\lambda_0)h_i = 0 $$ which implies that $h_i$ are not linearly independent. Thus in this case there are no roots in the region $\{\lambda \in \mathbb{R}^p\mid 1+\lambda'h_i>0\,,\forall i\}$.